i need help in this calculus question

Question

An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 2 cm, and the height is 10 cm.

Find the rate at which the water level is rising when the water level is 2 cm.

Answer 1

This is a matter of using similar triangles as well as the chain rule...

Note the diagonal of a 2 by 2 square is 2√2 whence the distance between the center of the square and any of its vertices is 1√2 = √2.

Now, a half diagonal base, altitude of the pyramid, and the slanted height of the pyramid form a right triangle. Any triangle with a base parallel to this triangle and a height that's a part of the pyramid's height is similar to this big triangle. So, let the smaller triangle have triangle height x, triangle base y.

By properties of similarity,

x/10 = y/√2 = [y•√2]/2 which implies that x/5 = y•√2

Also, by similarity, the smaller pyramid must have a square base with diagonal 2y and thus length y•√2 = x/5, and thus have volume

V = (1/3)(x/5)²(x) = (x³)/75 which implies that

dV/dt = [3(x²)/75] = [x²/25]dx/dt whence we get

45 = [(2²)/25]dx/dt which implies that dx/dt = 45•25/4 = 1125/4 = 281.25

Check to see if/where I messed up; it's late at night.

Answer 2

so the change in volume with respect to time dV/dt = 45 cm³/s

Volume of pyramid = 1/3*base*height = 1/3*s²*h

need to get rid of base in equation because we want to determine the rate of the water level rising (dh/dt)

using proportions in 2D: s/2 = h/10 where s is horizontal distance of water and h is height of of water at a given time. so s = h/5

plugging back in to the volume equation: V = 1/3*(h/5)²*h = (1/75)*h³

now we can take the derivative of the volume equation with respect to time:

dV/dt = (1/75)*3*h²*dh/dt

since we know dV/dt = 45 cm³/s and the problem asks for dh/dt when h = 2cm we can solve:

dh/dt = (dV/dt)*25/h^{2 =} (45 cm³/s)*25/(4cm²) = 281.25 cm/s